MHB Prove Euler's Equation for Functional $J(y)$

[evinda]

Hello! (Wave)

According to my notes, the following theorem holds:

If $y$ is a local extremum for the functional $J(y)= \int_a^b L(x,y,y') dx$ with $y \in C^2([a,b]), \ y(a)=y_0, \ y(b)=y_1$ then the extremum $y$ satisfies the ordinary differential equation of second order $L_y(x,y,y')- \frac{d}{dx}L_{y'}(x,y,y')=0$ (Euler's equation).

I want to prove that Euler's equation of the problem $J(y)=\int L(t,y,y') dt$ can be written in the form $L_t- \frac{d}{dt}(L-y' L_{y'})=0$.

Could you give me a hint how we could show this? (Thinking)

 

[GJA]

Hi evinda,

We compute via the chain rule:

$$
\begin{align*}
L_{t}-\frac{d}{dt}(L-y'L_{y'})&=L_{t}-\frac{d}{dt}L+\frac{d}{dt}y'L_{y'}\\
&=L_{t}-L_{t}-y'L_{y}-y''L_{y'}\ldots
\end{align*}$$

I haven't finished the above computation of $$\frac{d}{dt}y'L_{y'},$$ because I think you can probably get there yourself using the product and chain rules. If done correctly, you should be able to cancel a few terms and factor out a $$y'$$ term. From there, you will want to recognize that three of the remaining terms are precisely $$\frac{d}{dt}L_{y'}.$$ After noticing this, you can use the Euler-Lagrange equation to show that

$$L_{t}-\frac{d}{dt}(L-y'L_{y'})=0$$

Let me know if you're still stuck, or if anything is unclear/not quite right.

 

[evinda]

Isn't it as follows?

$$L_t-\frac{d}{dt}(L-y' L_{y'})=L_t-L_t+\frac{d}{dt}(y' L_{y'})=y''(t) L_{y'}(t,y,y')+y' \frac{d}{dt} L_{y'}(t,y,y')$$

If so, then how do we deduce that the latter is equal to $0$?

Or am I wrong? (Thinking)

 

[GJA]

Hi evinda,

I think the point of confusion is that we need to remember $y=y(t),$ so that when we write $L$, we really mean

$L=L(t,y(t),y'(t))$

With this in mind,

$\frac{d}{dt}L(t,y(t),y'(t))\neq L_{t}(t,y(t),y'(t)),$

because $\frac{d}{dt}$ is the so-called total derivative <--- which is basically a fancy way of saying the multivariable chain rule(see Total derivative - Wikipedia, the free encyclopedia for a reference)

I think an example to show why the previous equation doesn't hold may prove useful: Let's pretend that $L(t,y,y')$ is given by

$L(t,y,y') = t + y^{2}+3y'$

and that $y(t)$ is given by $y(t) = e^{t}$ (To be clear, I'm simply making up an example here, I am not claiming $L$ comes from a specific variational problem, nor am I claiming that $y$ solves the Euler-Lagrange equation).

Now, in this example, $L_{t}$ really means $L_{t}(t,y(t),y'(t)),$ which means to take the partial derivative of $L$ with respect to $t$ first, then plug in for $y(t)$. Doing so gives

$L_{t}(t,y(t),y'(t)) = 1$

On the other hand, the total-derivative (i.e. chain rule) $\frac{d}{dt}$ gives

$
\begin{align*}
\frac{d}{dt}L(t,y(t),y'(t))&=L_{t}(t,y(t),y'(t))\frac{dt}{dt}+L_{y}(t,y(t),y'(t))\frac{dy}{dt}+L_{y'}(t,y(t),y'(t))\frac{dy'}{dt}\\
&=1+2y(t)y'(t)+3y''(t)\\
&=1+2e^{2t}+3e^{t}
\end{align*}
$

From this example we see that $\frac{d}{dt}L\neq L_{t}.$ Does this example help shed some light on how the computation works out?

Returning to my original post, I began the computation as follows:

Hi evinda,

We compute via the chain rule:

$$
\begin{align*}
L_{t}-\frac{d}{dt}(L-y'L_{y'})&=L_{t}-\frac{d}{dt}L+\frac{d}{dt}y'L_{y'}\\
&=L_{t}-L_{t}-y'L_{y}-y''L_{y'}\ldots
\end{align*}$$

The $-L_{t}-y'L_{y}-y''L_{y'}$ terms in the above come from computing $-\frac{d}{dt}L.$ I did not finish the part of the computation involving the $\frac{d}{dt}y'L_{y'},$ which is what the ellipses represented. Do you think after reading this new post you can try to compute $\frac{d}{dt}y'L_{y'}$ using the product and chain rules again? If you're able to do this, my original post outlines what should happen from there. I'm sure you can get this, so I'm trying to avoid just giving you the full details right now. However, if you get stuck again, let me know and I will provide more help. Let me know if anything is unclear/not quite right.

 

[evinda]

Hi evinda,

I think the point of confusion is that we need to remember $y=y(t),$ so that when we write $L$, we really mean

$L=L(t,y(t),y'(t))$

With this in mind,

$\frac{d}{dt}L(t,y(t),y'(t))\neq L_{t}(t,y(t),y'(t)),$

because $\frac{d}{dt}$ is the so-called total derivative <--- which is basically a fancy way of saying the multivariable chain rule(see Total derivative - Wikipedia, the free encyclopedia for a reference)

I think an example to show why the previous equation doesn't hold may prove useful: Let's pretend that $L(t,y,y')$ is given by

$L(t,y,y') = t + y^{2}+3y'$

and that $y(t)$ is given by $y(t) = e^{t}$ (To be clear, I'm simply making up an example here, I am not claiming $L$ comes from a specific variational problem, nor am I claiming that $y$ solves the Euler-Lagrange equation).

Now, in this example, $L_{t}$ really means $L_{t}(t,y(t),y'(t)),$ which means to take the partial derivative of $L$ with respect to $t$ first, then plug in for $y(t)$. Doing so gives

$L_{t}(t,y(t),y'(t)) = 1$

On the other hand, the total-derivative (i.e. chain rule) $\frac{d}{dt}$ gives

$
\begin{align*}
\frac{d}{dt}L(t,y(t),y'(t))&=L_{t}(t,y(t),y'(t))\frac{dt}{dt}+L_{y}(t,y(t),y'(t))\frac{dy}{dt}+L_{y'}(t,y(t),y'(t))\frac{dy'}{dt}\\
&=1+2y(t)y'(t)+3y''(t)\\
&=1+2e^{2t}+3e^{t}
\end{align*}
$

From this example we see that $\frac{d}{dt}L\neq L_{t}.$ Does this example help shed some light on how the computation works out?

I think that I got it... (Thinking)

Returning to my original post, I began the computation as follows:
The $-L_{t}-y'L_{y}-y''L_{y'}$ terms in the above come from computing $-\frac{d}{dt}L.$ I did not finish the part of the computation involving the $\frac{d}{dt}y'L_{y'},$ which is what the ellipses represented. Do you think after reading this new post you can try to compute $\frac{d}{dt}y'L_{y'}$ using the product and chain rules again? If you're able to do this, my original post outlines what should happen from there. I'm sure you can get this, so I'm trying to avoid just giving you the full details right now. However, if you get stuck again, let me know and I will provide more help. Let me know if anything is unclear/not quite right.

So is it as follows?

$L_t-\frac{d}{dt}\left( L(t,y,y')-y' L_{y'}(t,y,y')\right)=L_t(t,y,y')-\frac{d}{dt} L(t,y,y')+\frac{d}{dt}(y' L_{y'}(t,y,y')) \\ =L_t(t,y,y')-L_t(t,y,y')-L_y(t,y,y')y'(t)-L_{y'}(t,y,y')y''+y''L_{y'}(t,y,y')+y' \left( L_{y't}+L_{y'y}y'+L_{y'y'}y'' \right)$Or have I done something wrong? (Thinking)

 

[GJA]

$L_t-\frac{d}{dt}\left( L(t,y,y')-y' L_{y'}(t,y,y')\right)=L_t(t,y,y')-\frac{d}{dt} L(t,y,y')+\frac{d}{dt}(y' L_{y'}(t,y,y')) \\ =L_t(t,y,y')-L_t(t,y,y')-L_y(t,y,y')y'(t)-L_{y'}(t,y,y')y''+y''L_{y'}(t,y,y')+y' \left( L_{y't}+L_{y'y}y'+L_{y'y'}y'' \right)$

Looks good!

 

[evinda]

Looks good!

Do we continue as follows?

$$L_t(t,y,y')-L_t(t,y,y')-L_y(t,y,y')y'(t)-L_{y'}(t,y,y')y''+y''L_{y'}(t,y,y')+y' \left( L_{y't}+L_{y'y}y'+L_{y'y'}y'' \right)=y'(t) (-L_y(t,y,y')+L_{y't}(t,y,y')+y' L_{y'y}+y'' L_{y'y'}) \\ =-y'(t) \left( L_{y}(t,y,y')\right)-\frac{\partial}{\partial{t}}L_{y'}(t,y,y') \frac{dt}{dt}-\frac{\partial}{\partial{y}}L_{y'}(t,y,y') \frac{dy}{dt}-\frac{\partial}{\partial{y'}}L_{y'} \frac{dy'}{dt}\\=-y'(t)(L_{y}(t,y,y')-\frac{d}{dt}L_{y'}(t,y,y')) \overset{\text{ Euler-Lagrange}}{=}0$$

Or have I done something wrong? (Thinking)

Also, have we shown in that way that the Euler equation of the problem $J(y)= \int L(t,y,y')dt$ can be written in the form $L_t-\frac{d}{dt} (L-y'L_{y'})=0$?
If so could you explain me why it is like that?
Because of the fact that we deduced that $L_t-\frac{d}{dt} (L-y'L_{y'})=0$ using the Euler-Lagrange equation?

 

[GJA]

Do we continue as follows?

$$L_t(t,y,y')-L_t(t,y,y')-L_y(t,y,y')y'(t)-L_{y'}(t,y,y')y''+y''L_{y'}(t,y,y')+y' \left( L_{y't}+L_{y'y}y'+L_{y'y'}y'' \right)=y'(t) (-L_y(t,y,y')+L_{y't}(t,y,y')+y' L_{y'y}+y'' L_{y'y'}) \\ =-y'(t) \left( L_{y}(t,y,y')\right)-\frac{\partial}{\partial{t}}L_{y'}(t,y,y') \frac{dt}{dt}-\frac{\partial}{\partial{y}}L_{y'}(t,y,y') \frac{dy}{dt}-\frac{\partial}{\partial{y'}}L_{y'} \frac{dy'}{dt}\\=-y'(t)(L_{y}(t,y,y')-\frac{d}{dt}L_{y'}(t,y,y')) \overset{\text{ Euler-Lagrange}}{=}0$$

Or have I done something wrong? (Thinking)

You got it! Nicely done :D

Also, have we shown in that way that the Euler equation of the problem $J(y)= \int L(t,y,y')dt$ can be written in the form $L_t-\frac{d}{dt} (L-y'L_{y'})=0$?
If so could you explain me why it is like that?
Because of the fact that we deduced that $L_t-\frac{d}{dt} (L-y'L_{y'})=0$ using the Euler-Lagrange equation?

We have actually shown this, because there are equal signs throughout. If you really want to convince yourself that the two equations are equivalent, you can START with the Euler-Lagrange equation

$L_{y}-\frac{d}{dt}L_{y'}=0$

multiply both sides by $-y'$ then follow the string of equations you wrote in your last post back to the form you want in the problem. Let me know if anything is unclear/not quite right.

 

[evinda]

You got it! Nicely done :D

(Smile)

We have actually shown this, because there are equal signs throughout. If you really want to convince yourself that the two equations are equivalent, you can START with the Euler-Lagrange equation

$L_{y}-\frac{d}{dt}L_{y'}=0$

multiply both sides by $-y'$ then follow the string of equations you wrote in your last post back to the form you want in the problem. Let me know if anything is unclear/not quite right.

So you mean that we could also justify that the Euler-Lagrange equation can be written in the form $L_t-\frac{d}{dt}(L-y' L_{y'})$ as follows, right?$$L_y-\frac{d}{dt}L_{y'}=0 \\ \Leftrightarrow -y' L_y+y' \frac{d}{dt}L_{y'}=0 \\ \Leftrightarrow -y'L_{y}+y'(L_{y't}+L_{y'y} y'+ L_{y'y'} y'')=0 \\ \Leftrightarrow L_t-L_t-y'L_y+y'L_{y't}+(y')^2 L_{y'y}+y' y'' L_{y'y'}=0 \\ \Leftrightarrow L_t-(L_t+y' L_y+y'' L_{y'})+y'' L_{y'}+y' L_{y't}+(y')^2 L_{y'y}+y'y'' L_{y'y'}=0 \\ \Leftrightarrow L_t-\frac{d}{dt}L+\frac{d}{dt} y'L_{y'}=0 \\ \Leftrightarrow L_t-\frac{d}{dt}(L-y'L_{y'})=0$$

Or have I done somethig wrong? (Thinking)

Thinking about it again, couldn't we also do it as previously and just start with the equality $ L_t-\frac{d}{dt}(L-y'L_{y'})=0$ and then use equivalences? (Thinking)